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Infinite square well

  1. Nov 4, 2016 #1
    1. The problem statement, all variables and given/known data
    A particle is in the n=1 state in an infinite square well of size L. What is the probability of finding the particle in the interval Δx = .006L at the point x = 3L/4?

    2. Relevant equations
    ψ(x) =√(2/L) sin(nπx/L)

    3. The attempt at a solution
    The problem states that because Δx is very small I don't need to do any integration. I've been integrating anyway because it's fun in this particular case- but my problem is what to do with x = 3L/4. I've integrated up to x= 0.006L in ψ2(x) and I'm getting the correct answer for if the particle was at x=L/2. What do I do here?
     
  2. jcsd
  3. Nov 4, 2016 #2

    BvU

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    You show your work and then we can comment :smile:. 'up to x = 0.006 L' makes me feel uneasy: up to x = 0.006 L from where ?
     
  4. Nov 4, 2016 #3
    So here is the integration from 0 to 0.006L. I tried the integration from 0.006L to 3L/4 also and im getting a number greater than 1...
     

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  5. Nov 4, 2016 #4
    Holy **** the answer was simply 0.006. can you please explain this to me?
     
  6. Nov 4, 2016 #5

    BvU

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    Disaster ! But the original exercise asks about probability to find the bugger in an interval of width 0.006 L at 3/L/4 , so I would be reassured if you integrated from 0.75 L to 0.7506 : or so. Since the sine varies only little in uch a small interval, you can just take the value. But, if I may ask: what do you integrate (what's the integrand) ? (edit: never mind, iI can see you do it right from the grey-on grey picture). Now about the value.
     
  7. Nov 4, 2016 #6
    I figured that since at L/2 there would be the highest probability at that point. At 3L/4 you're a little bit further right in the well and since the probability of finding a particle represents a sine wave then it would have to be a number smaller than what I got integrating from 0 to 0.006L (which was twice 0.006L) but how are you to know that it's at exactly 0.006L?
     
  8. Nov 4, 2016 #7

    BvU

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    Doesn't sound good to me, but I can be wrong....
    [edit] no, sounds good to me. Why ?
     
  9. Nov 4, 2016 #8
    The integrand is √(2/L) sin2 (nπx/L)
     
  10. Nov 4, 2016 #9

    BvU

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    You missed a factor 1/2: ##\quad{\bf 2} \sin^2x = 1-\cos 2x##

    [edit] and the numerics look mistaken too: for small x , ##\ x-1/2 \sin2x ## should be closer to 0.
     
  11. Nov 4, 2016 #10

    BvU

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    so at 3L/4 that sine squared is 1/2, and ## {2\over L } {1\over 2 } 0.006 L = 0.006 ##

    Simple exercise, lots to learn !. (not only you, me 2)
     
  12. Nov 4, 2016 #11
    I see. Thank you.
     
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